Results 1 to 11 of 11

Math Help - Relation, reflexive, symmetric and trasitive

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    50

    Relation, reflexive, symmetric and trasitive

    The given relation S on real numbers, R, is given by

    xSy <=> x*y > 0

    Determine wether S is reflexive, symmetric and transitive.

    How can I solve this assignment by proof or contradiction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Madspeter View Post
    The given relation S on real numbers, R, is given by

    xSy <=> x*y > 0

    Determine whether S is reflexive, symmetric and transitive.

    How can I solve this assignment by proof or contradiction?
    Note that reflexivity fails. Can you find the single element which this does not hold for? This is your contradiction.

    However, the other two do hold. Symmetry is pretty simple (it comes from the commutativity of * over \mathbb{R}). For transitivity, we can use the fact that x*y>0 \Leftrightarrow x, y > 0 \text{ or } x, y <0. Two cases:

    (1) xSy, ySz, x<0 \Rightarrow y<0 \Rightarrow z<0 \Rightarrow x*z>0 \Rightarrow xSz.

    (2) is identical, but we take x>0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    50
    The element which you speak of must be the fact that either x or y can be chosen as 0 - giving us a false statement. 0 is not bigger than 0.
    Is this how you mean reflexitvity fails?


    suppose in your 1) that x = 0 instead of x < 0 - wouldnt this give us a non-transitive relation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1486
    Awards
    1
    Quote Originally Posted by Madspeter View Post
    The element which you speak of must be the fact that either x or y can be chosen as 0 - giving us a false statement. 0 is not bigger than 0.
    Is this how you mean reflexitvity fails?

    wouldnt this give us a non-transitive relation?
    That is correct.

    Rethink the transitive part. If x~\&~y are related that is saying that they are not zero and both have the same sign.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    50
    Quote Originally Posted by Plato View Post
    That is correct.

    Rethink the transitive part. If x~\&~y are related that is saying that they are not zero and both have the same sign.
    of course

    if xSy <=> x*y > 0

    then either x or y can't be 0, cause they have to > 0, because they are related.

    ----

    this, however, does not hold for the reflexivity, where x can be 0.
    right?

    Do you think you have timefor one more task of relation? if not, have a nice evening and of course thanks for your help with xSy, it brought a bit more insight to relations.

    Here goes.

    The relation T on positive R+ = {x E R | x > 0 } is given by:

    xTy <=> [x E Q and x*y E Q]

    Determine reflexivity, symmetry and transivity. Use proof and contradiction.

    Im not sure wether to be confused by the extra amount of structere in this relation or to recognize this as very similar to the prior assignment.

    I'd go for not reflexive since either x or y can be 0

    The relation symmetric (commutitave rule, (x*y)=(y*x)

    though for transitive relation, we can't use your 1) and have to proof xRz by having x>0


    am I far off?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Madspeter View Post
    this, however, does not hold for the reflexivity, where x can be 0.
    right?
    No - the relation is both symmetric and transitive because 0 is related to nothing. If we have xSy then neither x nor y is 0 (as x*0=0 \forall x \in \mathbb{R}).

    As for your other problem, you should note that for x, y \in \mathbb{Q} \Rightarrow x*y \in \mathbb{Q}, and also that 0 \in \mathbb{Q}. You can use this result to show that your relation is, in fact, an equivalence relation.
    Last edited by Swlabr; September 27th 2009 at 11:12 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2009
    Posts
    50
    I need to clarify a couple of things.


    quote
    No - the relation is both reflexive and transitive because 0 is related to nothing. If we have then neither nor is 0 (as ).
    quote

    I thought you meant xSy = x*y > 0 failed to be reflexive?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1486
    Awards
    1
    I think the way you defined S is confusing.
    I read it to be a relation defined on \mathbb{R} as xSy if and only x\cdot y>0.
    In that case, the relation S is not reflexive, (0,0)\not\in S because 0\cdot 0 \not >0.
    Last edited by Plato; September 27th 2009 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Madspeter View Post
    I need to clarify a couple of things.


    quote
    No - the relation is both reflexive and transitive because 0 is related to nothing. If we have then neither nor is 0 (as ).
    quote

    I thought you meant xSy = x*y > 0 failed to be reflexive?
    That should read "is both symmetric and transitive", not reflexive. I've edited my previous post. I hope that stops any confusion...?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2009
    Posts
    50
    Quote Originally Posted by Plato View Post
    Deleted

    this is very clear!

    as goes for relation T:

    quote
    The relation T on positive R+ = {x E R | x > 0 } is given by:

    xTy <=> [x E Q and x*y E Q]
    quote


    Since x has to be bigger than 0, then my guess is that T is reflexive.
    x is related to x since any x*x > 0.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Sep 2009
    Posts
    50
    Quote Originally Posted by Swlabr View Post
    That should read "is both symmetric and transitive", not reflexive. I've edited my previous post. I hope that stops any confusion...?

    I see now , thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relation that is 1-1, reflexive, but not symmetric
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 16th 2011, 01:13 PM
  2. Transitive, Symmetric, Non-Reflexive Relation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 24th 2010, 07:12 AM
  3. Symmetric, transitive, not reflexive?
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 2nd 2010, 12:32 AM
  4. Reflexive, Transitive, Symmetric
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: May 20th 2009, 02:24 PM
  5. Reflexive and symmetric
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 12th 2009, 04:56 AM

Search Tags


/mathhelpforum @mathhelpforum